Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 80210 | Accepted: 17922 |
Description
We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.
Figure A Sample Input of Radar Installations
Input
The input is terminated by a line containing pair of zeros
Output
Sample Input
3 2 1 2 -3 1 2 1 1 2 0 2 0 0
Sample Output
Case 1: 2 Case 2: 1 解题思路:该题题意是为了求出能够覆盖所有岛屿的最小雷达数目,每个小岛对应x轴上的一个区间,在这个区间内的任何一个点放置雷达,则可以覆盖该小岛,区间范围的计 算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样,问题即转化为已知一定数量的区间,求最小数量的点,使得每个区间内斗至少存在一个点。每次迭代对于第一个区间,选 择最右边一个点, 因为它可以让较多区间得到满足, 如果不选择第一个区间最右一个点(选择前面的点), 那么把它换成最右的点之后,以前得到满足的区间, 现在仍然 得到满足, 所以第一个区间的最右一个点为贪婪选择, 选择该点之后, 将得到满足的区间删掉,进行下一步迭代, 直到结束。
#include
#include
#include
#include
using namespace std;
struct node{
double left;
double right;
}point[1005];
int cmp(node a,node b){
return a.leftd){
flag=1;
}
}
if(flag){
printf("Case %d: -1\n",cas++);
}
else{
sort(point,point+n,cmp);
temp=point[0].right;
for(int i=1;itemp){
sum++;
temp=point[i].right;
}
}
printf("Case %d: %d\n",cas++,sum);
}
}
}